# EXPERIMENT 4B

```Calorimetry: Latent Heat of Vaporization&copy;98
Experiment 6
Objective: To determine the heat of vaporization of liquid nitrogen.
DISCUSSION:
A calorimeter is a device that allows accurate determination of energy and
temperature changes during simple thermodynamic processes. A key requirement is that
it can act as an isolated, closed system. Many very sophisticated types of calorimeters
exist. One very common type is known to scientists as the Dewar flask, or simply
‘Dewar’. You know it as the Thermos bottle. An even simpler calorimeter is a
Styrofoam cup. It is a satisfactory isolated system for use in small experiments, such as
the one that is to be conducted here.
The science of making these measurements is referred to as calorimetry.
Experiments in calorimetry are quite common in high school physics and chemistry labs.
The prevalent experiments are these two: the specific heat, c, of an unknown metal; and
the latent heat of fusion (solidification), LF, of water into ice. Instead of duplicating those
standard experiments, we will attempt to try and find a more unusual constant: the latent
heat of the vaporization of nitrogen, Lv(nitrogen).
Let us review the standard process of heating ice from -20&deg; Celsius to superheated
steam at 120&deg; Celsius:
a. Heat the ice at - 20 to 0&deg; Celsius. This heating process involves no change of
phase, so the heat required is expressed by Q = mciΔT, where m is the mass of the
ice, ci is the specific heat of ice, and T is the temperature change of the sample.
The specific heat of ice is interestingly close to half of that of water. Ice has a
specific heat ci= 2,050 J/kg-K, while water has a specific heat of cw= 4,190 J/kgK. Recall that changes in temperature are the same on the Celsius and Kelvin
scales: ΔT is 20&deg;Celsius, which is the same as 20 Kelvin. So given a mass of ice,
m, the heat required to warm the ice is Q(1) is 41,000m.
b. Step 2 is the phase change from ice into water. The heat is expressed by Q(2)
=mL. We can look up L and find that it is 334,000 J/kg. So Q(2) is 334000m.
c. Heat water from 0 to 100&deg; C. So, using Q = mcΔT, we have Q(3) = 419,000m.
d. Another phase change. The water at 100&deg;C is turned into steam at 100&deg;C, and
given the latent heat of the vaporization of water, Lv is easy to calculate, Q(4) =mLv
=2,257,000m.
e. Heating the steam from 100&deg;C to 120&deg;C. Using steam’s c in this range,
2,000J/kg-K, we have Q(5) = mcΔT =40,000m.
6-1
The total heat required for the process is given by:
QTotal  Q(1)  Q( 2)  Q(3)  Q( 4)  Q(5)
QTotal  m41,000  334,000  419,000  2,257,000  40,000
So it will be easy to determine any thermodynamic process using the proper analysis of
the materials involved regarding phase, specific heat, temperature, etc. Let us now look
at the process we can use to find a fairly unusual constant, the latent heat of vaporization
of liquid nitrogen.
We will exploit the well known thermodynamic properties of water. We can use
hot water to vaporize liquid nitrogen. Thanks to the conservation of energy, we know
that the heat given off by the water is exactly the heat absorbed by the nitrogen. In other
words, the total Q of the system is constant. Instead of writing an expression of the type
Q lost by materials + Q gained by materials = 0, it is more traditional to state:
heat lost by hotter materials = heat gained by cooler materials .
For our process we have:
heat lost by hot water = heat used to vaporize nitrogen
Using the temperature change relation Q = mcΔT and Q = mL in the appropriate places
we have:
mwater cwater  T water  mnitrogen Lv ( nitrogen)
(1)
where ΔTwater = | Ti-Tf | represents the temperature change of the water.
EXERCISES:
1. Obtain an empty Styrofoam cup, and find its mass, mcup.
2. Take the cup to the teaching assistant who will add a small amount of liquid nitrogen
the mass of the liquid nitrogen and cup.
3. Subtract the mass of the cup to determine the mass of the remaining liquid nitrogen,
mnitrogen.
4. Take the second Styrofoam cup and obtain half a cup of hot water from the hot pot.
Now record it’s mass and temperature, Ti.
6-2
5. Immediately but slowly pour the hot water into the liquid nitrogen. The nitrogen will
boil rapidly. Be sure to hold the water above the escaping vapor while pouring
(explainWHY?).
6. Once the vapor has subsided, find the mass and final temperature (Tf ) of the water,
verifying that the nitrogen has completely evaporated.
7. Solve equation (1) for Lv(nitrogen).
8. Repeat the experiment two more times.
9. Find the percent difference between the standard value and each result as well as a
comparison with the average of the results. What is the benefit of performing the
experiment more than once?
6-3
Data Sheet – Experiment 6
Mass of
cup
mcup1
Mass of
nitrogen and
cup
m(LN2+cup1)
(kg)
(kg)
Mass of
nitrogen
Mass of cup
and water
Mass of
water
mnitrogen=
m(cup2+water)
mwater
m(LN2+cup1) mcup1
(kg)
(kg)
(kg)
Initial temp of
water
Final temp of
water
Ti
Tf
(&ordm;C)
(&ordm;C)
Latent heat of
vaporization for
nitrogen
Lv(nitrogen)
(J/kg)
6-4
```